Question: Solve the equation. $\dfrac{dy}{dx}=\dfrac{1}{6x-42}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\ln|6x|-\dfrac{x}{42}+ C$ (Choice B) B $y=\dfrac{\ln|6x|}{6}-\dfrac{x}{42}+ C$ (Choice C) C $y=\dfrac{\ln|x-7|}{6}+C$ (Choice D) D $y=\dfrac{\ln|x|}{6}+C$
Explanation: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{1}{6x-42} \\\\ \dfrac{dy}{dx}&=\dfrac{1}{6(x-7)} \\\\ 6\,dy&=\dfrac{1}{x-7}\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} 6\,dy&=\dfrac{1}{x-7}\,dx \\\\ \int 6\,dy&=\int \dfrac{1}{x-7}\,dx \\\\ 6y&=\ln|x-7|+C_1 \\\\ y&=\dfrac{\ln|x-7|+C_1}{6} \\\\ y&=\dfrac{\ln|x-7|}{6}+C \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\dfrac{\ln|x-7|}{6}+C$